I had an interesting phone call from a friend who wished to settle a Christmas family argument. The argument was over Deal Or No Deal. Imagine the case where there are only two boxes left closed, one belonging to the contestant and the other to someone else, and the million is in one of these two boxes. Now Noel offers the contestant to swap his box for the other, should he?
My friend was instantly reminded of the Monty Hall problem and said he should swap. His family disagreed, but he saw this of ignorance of the lessons of MH. So who was right?
My first inkling was that the difference between DOND and MH is that in MH the person who reveals the goat knows where it is. Therefore, they are choosing to reveal a certain piece of information, and this choice is important. But in this case the choice is random so does the analysis still stand? I think the easiest way to resolve this is to consider the conditional probability of the car/million being in the chosen box given that you are offered a swap.
Let M be the event a million is in the original box Let S be the event we are offered a swap For MH, we are always offered a swap: P(M|S) = P(M and S) / P(S) = (1/3) / 1 = 1/3 For DOND with N boxes, contestant always keeps choosing boxes until the million is revealed or only two boxes left. P(M|S) = P(M and S) / P(S) Clearly, P(M and S) = 1/N Since if the million is chosen, a swap will always be offered And P(S) = 1/N + (N-1/N)*(1/N-1) = 2/N i.e. the probability of a swap is the probability you chose the million plus the probability you didn't choose it and didn't open it in the random choices. So, P(M|S) = (1/N) / (2/N) = 1/2
So in the case of DOND the swap doesn’t matter. And just to check, here is the numerical simulation in python:
import random
size = 3
iters = 10000
choices = (['Goat'] * (size - 1)) + ['Car']
print "doors are", choices
os = 0
ss = 0
for n in range(0,iters):
random.shuffle(choices)
if 'Car' in choices[0:2]:
os += 1
if choices[1] == 'Car':
ss += 1
print "offered swap", float(os)/iters
print "should swap", float(ss)/iters
print "should swap | offered swap", float(ss)/os
edit: properly format code
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42 Comments
DAMM YOU AND YOUR LOGIC!!!
But I do concede that you have a 50/50 chance in DOND when you get to the final two boxes and the large cash prize is still up for grabs.
Best to just make a deal with the Banker then…
This is pretty stupid, you are answering the wrong question. There is no point working it out in terms of whether the swap is offered at that is random. Here is the interesting bit:
The problem for DOND is:
- You are down to the final 2 boxes
- the million is in one of them
- you are offered the swap.
Do you stick with the box you originally chose or do you take the swap? (Quite a dfficult question given the monty hall problem shows you should always swap)
The answer is, it doesn’t matter for deal or no deal because the other boxes are eliminated at RANDOM by the contestant. In the monty hall problem the wrong box(s) are eliminated by the INTELIGANT CHOICE of the quiz master, so you increase your chances if you swap.
Cheers
Tom Howard
I worked this through with Tom from a different angle. Which I think is valid and agrees with the conclusion that the swap has no MH like advantage:
Given that you are in the situation of one box left plus the one you chose originally and are being asked if you want to swap. The top prize is still in play. Oh and the usual 22 boxes at the start.
There are only two possible ways this situation could arrive:
Path 1. You chose the right box at the start
Path 2. You chose a ‘wrong’ box at the start and have now randomly eliminated 20 other ‘wrong’ boxes leaving the top prize box as the swap you are being offered.
Which gives the following:
Path 1.
Probability of choosing the top prize box at the start = 1/22
Path 2.
Probability of choosing any other box at start = 21/22
Probability of randomly eliminating all 20 other boxes and leaving the top prize box = (20/21)*(19/20)*…*(1/2) = 1/21
Total probability of swap box being top prize = (21/22)*(1/21) = 1/22
So both boxes are equally likely to contain the top prize.
As James said the difference with MH is that in DOND the boxes are randomly eliminated. If as in MH Noel knew where the top prize was and intelligently selected to eliminate 20 of the other boxes to arrive at the situation then MH would apply.
Tom, I’m not sure I follow your criticism. The point is I conditioned on whether the swap was offered. Conditional probability is pretty standard tool.
Your `proof’ is really just a claim as I see it.
When you say that the difference in this problem is that MH KNOWS where the prize is and in DOND it is random picks doesnt actually matter. The example STILL holds right but ONLY if you take ALL the other boxes as empty and ONLY want to win the 250,000. When you get to the 2 final boxes (an ‘empty one’ and 250,000) what has happened is that you have played out the role of MH BY PURE CHANCE. Hence you should swap. In any other given situation with 2 boxes in DOND swapping will make no difference as there is always a prize on offer. I am sure this is right!
Dannie,
Your conclusion is fundamentally flawed. You say that the probability it is in your box is 1/22 and the probability it is in the sole remaining box is 1/22. What happens to the other 20/22 chances?……….
Part of the problem with this discussion is that the terms of reference are not clearly defined. The question, I believe, should be: “in DOND when there are two boxes left one of which contains nothing (or 1p) and the other contains £250,000, what is the probability that the box you chose contains the £250,000?”
The answer to that has to be 1/22. You chose it at random from 22 boxes. Nothing has intervened. The probability remains the same. Therefore it must be that the probability that it is in the other box is 21/22.
The key to the question is in the fact that you choose the box at the beginning of the game when there are 22 boxes.
Ollie is right that it is the same as MH albeit by pure chance and whether you have the knowledge that you’re taking out ‘empty’ ones or this happens by pure chance that you take out ‘empty’ ones makes no difference.
I think you are all being misdirected…
The salient point here is that whether swapping the box or not, Noels house party will never again disgrace our screens.
Surely that makes everyone winners, whatever particular box they are perving over in the meantime.
Hirsty
Think of it that YOU eliminate 20 boxes, you gain no other information along the way. You actually pick one, eliminate 20 and are left with one more.
This is no different from picking just two boxes at the outset (other than Noel would not have a 1 hour show to fill).
So each box is 1/22 and thus a 50/50 probability.
Everyone appears to have mixed opinions here.. so I’m going to give a scenario -
You pick a box at the beginning of the game, 1/22. Let’s say you eliminate boxes one by one and come down to two boxes. Obviously one box contains a larger sum than the other. Statistically, should you swap or stay with the original box. Which, if either, option has more of a chance of providing the larger sum?
Okay, so I’ve been turning this in my head over and over. RE: Dannie’s calculation, you have:
Path 1.
Probability of choosing the top prize box at the start = 1/22
Path 2.
Probability of choosing any other box at start = 21/22
Probability of randomly eliminating all 20 other boxes and leaving the top prize box = (20/21)*(19/20)*…*(1/2) = 1/21
Total probability of swap box being top prize = (21/22)*(1/21) = 1/22
Now, we’ve all been assuming that you’re down to a scenario where you’ve successfully eliminated 20 “empty boxes.” In DOND, you have to do it randomly, in MH, it’s by intelligent choice. This does not matter.
If we are to assume that you successfully eliminate 20 boxes, and are left with your box and another, and one of those boxes has the prize and the other is empty, you SHOULD SWAP. Why?
Because you have a 1/21 chance of going down path 1, but a 21/22 chance of going down path 2. Remember, we’ve already stated that you’re in a scenario where we happen to be able to eliminate 20 empty boxes. That is, we KNOW we are on path 2! Therefore, to win, you should swap!!!
^Okay, let me clarify it a little. We don’t KNOW we’re on path 2, but we KNOW we are comparing those two paths, because we’ve always talked about a scenario where there are only two cases left, and one has the prize money.
So if we know that, let probability of getting down to two cases where one has the money = 100%.
Prob(selecting the prize case & getting down to two cases) = 1/21
Prob(selecting an empty case & getting down to two cases) = 21/22
Everyone here seems to be missing a key thing here. Let’s say that at the end of the game you’re left with 2 boxes, the top prize and the bottom prize. You can, of course, apply the reasoning that the odds were 1/22 for you to pick the top prize, having opened all the boxes but the remaining box you can say that it is therefore 21/22 that the other box contains the jackpot. However you could just as easily turn this reasoning on it’s head. For example, at the beginning it was also 1/22 that you chose the bottom prize, and when you reach the end of the game, according to the exact same conditional probability some of you are employing, it’s also a chance of 21/22 that the last remaining box contains the booby prize. This is obviously a paradox, hence the reasoning used to get here is false.
The correct solution is to consider the problem the following way. Let’s imagine you get to the final two boxes again, 1 box contains the jackpot and 1 contains another prize. There are 2 ways you could have got here, either by choosing the jackpot as your box, or by choosing to open every box but the jackpot.
The odds of choosing the jackpot are 1/22
The odds of leaving the jackpot until last are 21/22*20/21….*2/1 = 1/22
Hence this situation will occur equally often via both routes, thus 50% of the time you will have the jackpot and 50% of the time you won’t. It’s a coin toss.
If you use the pre tag, perhaps around code tags just for fun, Wordpress (actually, the web browsers) will not lose your indents.
This argument has been worked to death on a probability newsgroup – I used to be fooled into the MH analogy until I realised that the chance of your picking the £250K was 1/22 ONLY at the very begining – each time a non – £250k box is opened ( at random ) the odds shorten – until
ultimately you arrive at 50/50.
Swapping makes no difference to your probability of success.
Well… I’ve certainly enjoyed all these arguments for and against the equivalence of DOND (Deal or No Deal) and MH (Monty Hall), but let us look at the weightings of choice based on the values of boxes remaining and does that affect the choice. Strictly speaking, no… as it would appear that you always have a 50% chance of choosing a higher or lower box.
Nonetheless, I would like to see a probability tree of all the outcomes, with the option to swap at the end… and I think that would resolve it for all scenarios.
FINAL POINT: If it’s different to MH then with 50% probability, there’s no harm in swapping. If it’s the same as MH then swapping is better. As we have split opinions, we should swap anyway (even if correct in the opinion that MH is different)… Therefore I conclude that you should always swap!!!
I was just watching DOD and the scenario unfolded where the guy had three cases: $100, $10,000 & $750,000.
He took the deal, but when asked what he would done he said Case 3 ($10,000). He then said he would swap (which I thought he should of bc of the MH problem). Had he swapped he would have won the $750k.
I was trying to explain why he should have swapped to my roommate, but I’m not much smart in the maths, so came to the Internet. I was interested to see such debate and to see the consensus is a no swap.
I disagree and say they should swap.
Think of it this way:
There are 26 cases. You obv. have 1-26 of picking the $1M. But let’s not limit it just to that. Really, there are about 5 Big Money cases. So let’s say those are the cases you are after. The rest are Crap Cases
Your odds are 5-26 or picking a Big Money Case, and 21-26 of picking a Crap Case.
So from the start, the odds are far likelier that you picked a Crap Case vs. a Big Money Case.
If you have gotten down to 2 cases, one we know is a Crap Case and the other a Big Money Case, it doesn’t matter if it was revealed to you intentionally or by random, you are still at that same point. It is irrelevant how you got there.
I know the perception is it is 50/50 between the Crap vs. Big $ case because you see the two in front of you, but I refuse to believe it.
When you picked the cases there was a 21/26 chance you picked a Crap Case. Thus, when given the option, you should swap your Crap Case with what is likely to be a Big $ Case. You have now inversed your odds of being 5/26 that you now have a Crap Case. i.e. 21/26 of having a Big $ Case.
Again, I am not a statistician by any means, but this sounds pretty logical to me, and in spirit with the MH problem, which everyone seems to think doesn’t hold water in this case. (No pun intended.)
Tom, you really haven’t thought that through. The probability that the OTHER box started with a big one is ALSO 5/26. Your own (flawed) logic could equally be applied from the perspective of the other box, to say you shouldn’t swap.
The odds do NOT remain the same as the start because every box that is opened change the odds of what is in your box and every other box. According to your logic, you would still have a 1/26 chance of having the sum that is in a box that has already been opened. Clearly, once a box has been opened, the probability that you chose that box at the start is no longer in existence, hence the probabilities continue to evolve rather than remaining consistent with the initial probabilities.
Hey idiots. Swap the case. Monty Hall DOES apply because you are given knowledge throughout the game – just as if Monty Hall was telling you a goat was inside of each case eliminated. It doesn’t matter if Monty Hall tells you this, or if you just get lucky on DOND. The fact remains that after your choice is made, you are given knowledge that $1 million still remains. This is new knowledge. By the time you get down to the end with only two cases remaining, you KNOW that one of the cases has $1 million, and the other case does not. This knowledge could have been given to you by Monty Hall, Helen Keller, Jack Benny, or by your stupid luck. Nonetheless, you are left with two cases and one has $1 million. You know that one of them has $1 million. If the total number of cases is 22, then the odds that your case has $1 million is 1 in 22 (these are the original odds when you picked your case). The odds that the remaining case has $1 million is 21 in 22.
Switch cases. The Monty Hall principle does apply.
It is a completely different story if Howie did not tell you that you were left with two cases, one with $1 million and the other with something else. Pretend that you started with 22 cases and just randomly eliminated 20 cases. The big information board isn’t working, and hot chicks holding the cases never open the cases. They just sit there looking stupid and so damn hot, and they forget to open the cases to show you what you just eliminated. So, you have just eliminated 20 cases. You don’t know what you eliminated. Your thumb is up your ass and you are now left with two cases. Should you switch? You should switch only if it gives you the opportunity to walk up to the hot model holding the case. It does nothing for your odds of winning. But, hopefully you were able to get a little sniff of her perfume. Odds are you already lost the $1 million with one of your earlier eliminations anyway. You don’t know anything about the last two remaining cases. If you switch or not, the odds are still 1 in 22 that you picked $1 million.
Take that stupid probability math and stick it where the sun doesn’t shine. It’s so pretentious. Dont’ let someone confuse you with flawed logic even if they use fancy numbers.
Thanks, John for your analysis, and for the show Cops
That ‘pretentious’ ’stupid probability math’ is backed up by a simulation that requires no probability maths at all. Have you found a flaw in that as well?
Thank you James Casbon. Now I’m the idiot because I thought I knew what I was talking about. I apologize for my abrasive language.
I’m back to believeing it’s a 50/50 split when we get down to the final two cases. In my scenario I assumed the $1 million took the path down to the final two cases. But, just as easily, the goat could have taken that path too. In the end, I am left with two cases that took a path. I don’t know which scenario happened.
Monty Hall knows which scenario happens. He eliminated the goat for you. So, in 21 of the examples, you are going to be presented with an ending scenario that involves $1 million and a goat. But, you really don’t desirve this because on DNOD, you would have arleady eliminated the $1 million earlier. You woudl never get to a Monty Hall scenario at the end. The only time you do get to the end of having $1 million in a case and a goat int he other case is the 50/50 case.
I’m not an Einstein. But, Einstein proved all of his theories through logic and reasoning. I don’t like numbers because I work with them as part of my job. I know how they can get manipulated. I value companies. OMG. You just wouldn’t believe the things that people do with numbers when it comes to valuing a company….
If I could delete my comment sent yesterday, I would. Not becuase I was wrong. But, because I called all of you nice people idiots.
Its a problem I began thinking about last night after watching the movie ‘21′ which included the MH problem.
In the end I wasn’t sure, on the one hand the losing doors are opened on purpose and on the other they are opened randomly and luckily. Does this change probability? I basically went back and forth with my reasoning thinking it could be 50/50 or maybe 25/26(US version of the game).
I had settled on that it was advantageous to switch. When you pick a case you have 1/26 chance of hitting the million and the odds of not selecting the million being 25/26. Which means that the odds of the million being in what of the ladies cases is 25/26. As cases are eliminated these odds shouldn’t change.
I do however understand Johns last point about the Goat having similar odds, so by tomorrow I may change my mind :p. But if you look at it as 1 car vs 25 goats, does the goat really have similar odds?
Suppose you pick your case, and then you pick the last remaining case. Both of these have the same probability of being the top prize (1/26; US version). Do you agree?
Yes, okay, then, you go on and open 24 cases. If you’re lucky to never reveal the top prize, what are the probabilities of the two remaining cases? You’ve already agreed they’re the same.
What makes this different from MH is that in MH you always reach the opportunity to swap. In DoND, it’s highly unlikely to get to two final cases and still have the top prize left.
This could also be simulated with a Python program; you’ll have to throw out a bunch of scenarios until you get to a scenario where one of the last two has the top prize. About 500 times out of 1000, you will have picked the top prize initially.
import random
iters = 1000
random.seed()
million = 1000000
count = 0
picked = 0
while (count < iters):
cases = [0.01,1,5,10,25,50,75,100,200,300,400,500,750,\
1000,5000,10000,25000,75000,100000,200000,300000,\
400000,500000,750000,million]
random.shuffle(cases)
yours = random.choice(cases)
del cases[cases.index(yours)]
for i in range(len(cases)-1):
pick = random.choice(cases)
del cases[cases.index(pick)]
if yours == million or million in cases:
count = count + 1
if yours == million:
picked = picked + 1
print “out of %s times, your initial pick was top %s times” % (count,picked)
You can’t change your probability of choosing the winning box based solely on the knowledge you have obtained through the game, as your sample set doesn’t change as you go, you are just seeing it. From the beginning, if there are 26 boxes from which to choose and only one winner, you have a 1/26 chance of winning. So if you are so fortunate as to have seen all of the others but one, you need to swap boxes. Even when you get to the end, you are still looking at the same initial conditions and the same sample set (the money didn’t change boxes during the game), but it just happens that you have seen 24 other boxes.
This idea of a non-changing sample set is why the Monty Hall problem works out as it does. It has nothing to do with the host or the fact that the host knows where the winning box is. If Howie and “The Banker” both know where the winning box is, as long as Howie can bluff it and “The Banker” makes his side offers in a standard way, the game is still “fair.”
Chris I will very simply show you why that logic is wrong and swapping makes no difference.
You are left with two cases, one wins $1m and the other $1. Now apply all the logic you and some other have done but every time you use the words “winner” or “$1m box” exchange that with “$1 box” and you will see the error. The EXACT same logic can be applied for the path of the 1$ box. If you are in the land of backwards where $1 was actually more money than $1m then according to your logic then you should swap boxes in order to win the $1 case. They can’t both be right hence the paradox.
Your 2nd paragraph is also wrong. The host knowing where the winning box is is precisely what makes it different.
If you have 3 boxes and only 1 wins and you pick a random box, the host will remove one of the losing boxes from play but he will never remove the winning box from play and that is where you gain the equity in swapping.
What if in the MH situation, you were given an opportunity of choosing up one of 10 cases. Then Monty eliminated the other choices for you one at a time, knowing that he only eliminates the wrong case. Between each elimination, you are given the option of changing cases.
He keeps doing this until there are only 2 cases left, should you switch?
Look, if you think there’s a difference between these two problems, it stems from the fact that you do not understand the veridical paradox of the Monty Hall Problem.
It does not matter at all whether Monty Hall or Howie Mandell, or Richard Nixon knows where the “winning” box is. The fact is that at the outset of the problem you are presented with a number of choices, say 25. One of those choices– whether you chose it or someone else assigns it to you– is set aside as “yours”. What the contestant is asked to decide is whether or not he selected– or was assigned– the “winning” choice to begin with. In either the case of “Let’s Make a Deal” or “Deal or No Deal” The long odds are against that initial choice. This is because, as any decent explanation of the MHP will illustrate, you are choosing between a 1/25 shot (your case or door) or a 24/25 shot (all other cases or doors).
It makes no sense to offer further explanation of the MHP as it is illustrated in detail here: http://en.wikipedia.org/wiki/Monty_hall_problem
It also doesn’t matter what any of you think about whether or not this problem works one way or the other. There are well established proofs. As in the MHP, you increase your chances of winning by trading your case for the one remaining case in DOND. If you don’t understand it, you don’t understand it– that doesn’t change the outcome.
And by the way Tim, you’re a dummy. Monty Hall or Howie Mandell or the Banker aren’t selecting for the $1 case, they’re selecting for the million dollar case, so of course the two are not interchangeable.
I tried to recreate this problem practically, but I used 21 goats and a car. My results were as follows:
I ran over 21 goats.
Therefore always pick the car.
Nuff said…
I believe there is a drastic flaw in every analysis given above. I will state my case, but will not take it personally and in fact be very pleased if someone gave me a counter argument and called me a fool.
i do in fact believe that the monty hall problem applies to deal or no deal, but not as we have analysed it.
in the monty hall scenario, we ‘choose’ a door. give this door property C (please excuse my crude logic), such that it was chosen from the total of 3 doors. we know that at least one goat (or empty door) must remain amongst the two, which is then shown to us. let us give the remaining door the property R, such that it remains after the ellimination of one of the goats. we all know the probabilities here or swapping or sticking.
in the deal or no deal scenario, the above analyses are treating the box which we carry to the table as we did door C – that it was chosen from the list of boxes. in fact it is NOT the same. this box was given to the contestant randomly, and it is not subject to the opening and elliminating that the other boxes must survive to be one of the remaining two.
i suggest that the box with the property C – being chosen from the entire set – is the box that remains unopened at the end. we actually chose not to elliminate this box, even though we could have, therefore THAT is the box with the high probability of not being the big money.
key point: in MH, there is one door which becomes ’safe’ from ellimination – that is door C. this door has a 2/3 chance of containing a goat/being empty. in deal or no deal, there is also one box which is ’safe’ from ellimination: the one you choose (although randomly, unlike MH) not to elliminate! therefore THAT box is box C, not the one given to you at the start. in MH, the only door which is unelliminatable at the start is the car, whether you chose it or not – the only box expressing this quality is the box you are given to bring to the table. in MH, you choose a box, then one of the other boxes is elliminated, revealing one. in deal or no deal, you are TOLD which box will not be elliminated, then you (ultimately) choose one of the remaining 21 to be left unopened – THIS is our C box.
so, IF (and i stress IF) the MH problem applies here, then you should not in fact swap, but stick! if you arrive at the situation of having 2 boxes, one with 250,000 and one with something else, the only ‘choosing’ that has occured is by you (as in MH) choosing not to elliminate that box. in deal or no deal, YOU also play the role of monty hall. you are not able to elliminate the box brought to the table, as monty hall is not able to elliminate the car (if it is in fact among the two remaining boxes, which is 2/3). but in terms of ‘choosing’ the correct box, that box is the one left with another person, not yours. the boxes you elliminate along the way to this scenario play the role of the revealed ‘at least one goat remaining’ door in MH, and as the box you were given randomly has is more likely to have been the big money since it was always safe from ellimination, but the other box was not.
Hi Guys
Deal or No Deal and the Monty Hall thingy are not the same scenario, here’s why…
DOND: There are only two processes by which you can reach a big box / other box scenario, these are i) pick the big box first up or ii) pick every other box until there’s only the big box left. As these are the two extremes of the selection sequence, the likelihood of picking the big box first or picking the big box last are the same e.g. picking the big box first is 1:26, selecting the last box by choosing every other box first is 1:26. As the likelihood of picking the big box as your first choice or your last is the same, changing does not improve your chance of winning.
Monty Hall: you have three options, one winner, two losers, 2 times out of 3 you’ll not choose the winner when you choose. Where the skew occurs is that at this stage, there is a no lose scenario as a losing option is then eliminated with the proviso that the box you’ve chosen can’t be removed. So, having been left with two boxes, the inital chance of you making an incorrect choice at stage one hasn’t changed making it more likely that you have a losing box, so in this problem you should change.
This is a slight deviation, but I think it’s quite interesting. The only way to replicate a MH scenario within DOND is to have selected the winner as your first box, as this is the only scenario whereby you can get to a three box situation with a guarantee of removing a loser. This is, as you’ll all have grasped instantly, the only MH scenario whereby swapping loses.
http://en.wikipedia.org/wiki/Monty_hall_problem#N_doors
Explanation is provided here…
In my own words:
The difference between MH and DOND is that in MH, the odds of you picking the right door is 1/3 while picking a goat is 2/3. The host opening a door with a goat does nothing rule out any scenarios/cases because whether or not you picked the right door he will open a door with a goat, and whether or not you picked the right door there is always a door with a goat for the host to open and he will never open the door with the car. The set space remains the same as in all cases the host will open the door with a goat. Thus you should swap.
In DOND, the odds of you picking the right door is 1/22; the first time you open a box (e.g box A) there is 1/22 that box A is the million dollar one; if it’s not, the set space is now smaller as you can now eliminate the case where box A contains a million dollars. then the remaining 21 boxes now each have 1/21 chance of being the box containing the million dollars.
With each selection of an empty box the remaining boxes have an increasing chance of containing the million dollars, until two boxes are left, namely your box and the remaining box out there. The odds are then 1/2.
The Monty Hall problem definitely doesn’t apply. Consider if you tried to play Let’s Make a Deal in the same style that Deal or No Deal is played.
If your initial choice is a car, any other door you pick will reveal a goat. If your initial choice is a goat you, the contestant, must then randomly pick one of the other two doors to eliminate. If you reveal the car it’s game over, and the situation no longer applies to our debate anyway.
The crux of the problem is… there’s no way to tell if your initial choice was the car or if your initial choice was a goat and you just got lucky and didn’t reveal the car. The only way to know for sure that your initial choice was a goat is to reveal the car, but you lose anyway at that point.
I suppose some of you are still clinging to the Monty Hall Problem though. Your explanation will be that there is 1/3 chance to pick the car. Swapping would give you the goat those times. There is 2/3 chance that you pick the goat. Swapping would give you the car those times. So from the outset, you have 2/3 chance to win by swapping, right?
No, WRONG. Because if you picked a goat first, remember that YOU have to randomly eliminate the other door. Half the time you will eliminate a goat, but on the OTHER half you will eliminate the car (in which case even if the game did go on, if you swapped you would receive a goat, obviously).
We can break this down further for people who still don’t understand. The items in brackets are the chosen doors.
Initial doors . . . . . what’s left
[G1] .G2. .Cr. –> [G1] & .Cr. or LOSE
.G1. [G2] .Cr. –> [G2] & .Cr. or LOSE
.G1. .G2. [Cr] –> .G1. & [Cr] or .G2. & [Cr]
As you can see there are 2/6 instances where we have a goat first, and swapping gives us the car. And the other 2/6 instances show us having a car first, but swapping to a goat. And if you decide to stay, 2/6 of the time you initially have the car, and 2/6 times you initially have a goat. Thus, it doesn’t matter whether or not you swap because no matter what, it’s 50/50. You might also note that if Let’s Make a Deal was played in this manner, there would be a 1/3 chance to win from the outset, and a 1/2 chance to win if you managed to eliminate a goat. Hope this clears it up so there won’t be any more arguing.
“No, WRONG. Because if you picked a goat first, remember that YOU have to randomly eliminate the other door. Half the time you will eliminate a goat, but on the OTHER half you will eliminate the car (in which case even if the game did go on, if you swapped you would receive a goat, obviously).”
I have to dissagree with you Chris. (I think!)
The comparison only applies when you get to the end and you have a swap choice between 2 boxes. In this case you do not have to take a chance on elimination as it has already happened.
Guy
I think we’re all forgetting the real-world here.
Would a TV show format which is sold all over the world give unfair odds, even in unlikely situations? I find it hard to believe that a game show based on pure luck would do that.. the television companies would surely lose more money that way.. even if none of the contestants knew of/applied the MH idea, they’d still lose more money that way than by giving it an even chance.
However, I’m still unconvinced mathematically one way or the other.
MH is different to DOND, and most of the analyses above agreeing with that have given good reasons. Here’s a another simple way to look at it in terms of information or surprise value.
That Monty can reveal a goat tells you nothing. You already know there’s at least one goat there somewhere. He opens a door and shows you one. No surprise. You learn nothing. Your odds of having picked the right door stay at 1/3.
In DONR every time you open a box randomly and it isn’t the top prize you are relieved. You are surprised. Why? Because you have gained information and your odds adjust accordingly.
After much head-scratching, I now think swapping boxes makes no difference to odds of winning the top prize in DOND, as there is an important difference between DOND and MH (In DOND, 2-box offer of swap occurs only when 20 other possible outcomes have NOT occurred – so this affects the odds)
MH –
1 in 3 times – car selected, Goat reveal, 2 options : SWAP = Goat NO SWAP = CAR
2 in 3 times – goat selected, Goat reveal, 2 Options : SWAP = CAR No SWAP = Goat
Therefore SWAP is twice as likely to get the car than no swap.
DOND –
1 in 22 times Top prize selected first
20 in 22 times Top prize not selected first but eliminated before only 2 boxes are left.
1 in 22 times Top prize left till last
So in 2-box finish, the middle 20 chances are ruled out and the 1 in 22 chances now both become 1 in 2. Therefore, swapping makes no difference.
If in DOND 20 non-top prize boxes were taken out by a host who knew where the top prize was, it WOULD be favourable to swap boxes as MH principle would apply.
DOND played like MH –
1 in 22 times Top prize selected first
21 in 22 times Top prize in last box [20 options removed by intelligent selection]
Therefore HIGHLY favourable (21 times) to swap in this hypothetical game (like MH)
To sum up, you’re most likely to NOT choose top prize at first. Swapping at this point makes no difference. Any information which reduces the number of non-top prize options in remaining set of options means this groups is more likely to contain the top prize.
The 2-box DOND offer when Top prize left is unlikely to happen – but equally likely to happen because of original choice being Top prize or Top prize not yet selected or opened. Hence equal probabilities in this situation.
Firstly, I don’t think that randomly pick boxes until the two are left in DOND makes any differences to the favourability of a swap. For example, take the monty hall problem, if monty does not know where the car is and choses a door to reveal at random he may pick the car or one of the goats. However if he does choose a goat, the final situation is the same as the original problem, you know it is more likely you have picked a goat and as the car still remains you should swap. The fact that he randomly choose only means it is less likey to reach this final scenario but does not alter the probabilities any further than that. If he had revealed the car the game ends anyway and the problem does not apply. Think of it like rolling a dice, the previous rolls don’t affect the next roll, you are just as likely to roll a 6, even if you had rolled a 100 in a row before that.
Therefore in DOND if you say there are 21 boxes which are failures and 1 sucess it is modelled as a binomial, a fixed probability at the start and either a success or failure outcome. Let us say the success is the 250,000 prize. you have a 1/22 chance of picking it at the start, you then remove 20 more boxes and are told none of them contain the success. so you are left with the box you picked, which has a 21/22 chance of being a failure, and another box. As you almost certainly didn’t pick the 250,000 at the start, statistically it is probably not in your box hence you should swap.
The Monty Hall problem applies only to the situation given above, which is very unlikely to occur, because it involves choosing 20 boxes in a row which do not contain the success, but when it does occur you know to swap. As you have defined all other boxes as failures if you end up opening a box with the 250,000 (the success box) then you have lost as you are left with only failures which you cannot differentiate between. As DOND has varying prizes, not a simple win or lose scenario you can only apply the monty hall if you class 21 of the 22 boxes as failures. If you end up having to choose between two boxes which are known not to be the success then you are equally as likely to get a higher prize if you swap or not. The Binomial model no longer applies.
I hope this makes sense. I have tried to reason using logic and common sense but a mathematical proof is obviously more favourable.
I think I can prove that MH does NOT work in DOND…
In the same way that MH is scalable upwards (see wikipedia entry), DOND can be condensed to what happens when 3 (or fewer) boxes are left in play.
If you had space & time, you could create the same list of outcomes starting with 22 or more boxes if you like (you’d have a lot of iterations where desired winning box is eliminated early…) – the odds don’t change once you’re down to 2 boxes…
I’ve tried to illustrate this below, assuming M to be the value (or car) you want to win, and A & B to be lesser (or ‘goat’) values.
Let’s assume you’re down to the last 3 boxes, about to ‘remove’ one, in the same way as MH would ‘identify’ one of the goats.
You’re in one of 3 situations at this point – you either chose M (the high value box) at the outset, or you chose a goat value (A or B) – this is the left hand column below, with your selected box in brackets. The top 1-4 scenarios illustrate what happens if you brought (M) to the table – in 1 & 2, you remove box A from the game, and are left with (M) and B – you then can keep (M) for a win, or swap to B and lose.
In 3 & 4, you remove B from the game, leaving (M) and A – keeping (M) wins, swapping to A is a loss.
5-8 illustate what happens if you started with (A), and have M still in play with 3 boxes to go…
9 -12 illustrate what happens if you started with (B), and have M still in play with 3 boxes to go.
In the first 4 examples, this relates to what happens in MH when you picked the car door first.
Monty shows you a goat, and if you swap, you lose – this is your 1 in 3 odds-of-winning equivalent.
In the examples 5-12, this is where, in MH, you initially chose a goat – MH can then show you the other goat, and any swap here would give an increased chance of a win (2 out of 3 times).
Here’s where the fault lies in the comparison, and why MH is not compatible with DOND – in scenarios 5-12, you as the player, can remove the M value box from the game (specifically 5,6,9 & 10). These scenarios don’t occur in MH – it would be the same as MH not only showing you, but also removing the car – this doesn’t happen in MH, and it’s why you’re always left with a chance to win.
Even if you removed any iterations from DOND where the M box isn’t present in the final 2 box choice, you’ve still got 8 potential outcomes where you can chose between M and the remaining A or B box (asterisked *). 4 of these outcomes can win (**) – two if you swap, and two if you keep – and 2 chances in 4 is 1/2 or 50/50 – therefore a SWAP GAINS NO ADVANTAGE!
In other words, however you get to the point, at two boxes, where the high value M box is still in play (either at the table, or because it’s left in play) there’s the same chance that a swap will result in a win, as there is a loss…
1. (M) A B -> (M) B = (M) no swap wins**
2. (M) A B -> (M) B = swap B loses*
3. (M) A B -> (M) A = (M) no swap wins**
4. (M) A B -> (M) A = swap A loses*
5. (A) B M -> (A) B = no swap loses
6. (A) B M -> (A) B = swap loses
7. (A) B M -> (A) M = (A) no swap loses*
8. (A) B M -> (A) M = swap M wins**
9. A (B) M -> A (B) = (B) no swap loses
10. A (B) M -> A (B) = swap A loses
11. A (B) M -> (B) M = (B) no swap loses*
12. A (B) M -> (B) M = swap M wins**
ok, I’m a little confused by some of the long maths proofs, which may or may not work, I have no idea
However, I am convinced that swapping is 50/50 for the following reason.
If you knew that every game you’d get to a situation with £250,000 and the rubbish, then clearly you should swap, aka the monty hall, and win 21/22 times. However, you don’t know this, and in fact, if you have picked the £250,000 box to start with, you are 21 times more likely to reach this situation than if you hadn’t picked it, which makes it kind of equal. I’ll try and explain why below.
So imagine a situation where you play 22 games. In 21 of those games, you don’t bring £250,000 to the table. Now the chances of getting to this final ’showdown’ are 1/21. So from the start of the game, the chances of getting to the showdown without bringing £250,000 to the table is 21/22 * 1/21 = 1/22. However, if you do bring £250,000 to the table, you are guaranteed to reach this showdown, so the chances of reaching the showdown bringing £250,000 to the table is 1/22. (For those who have made comments earlier about these not adding to 1, that’s because the specified showdown only happens 1/11 times). Hence, as the chances of reaching this showdown are equal, regardless of whether the £250,000 box has been brought to the table, neither is more likely than the others.
By the way, I’m sure this is adapted from other people’s comments
I came to this page earlier today fully believing that you should swap …
Ok, this is the way I see it. Let’s call the box that you pick at the start, the one that stays with you throughout the game, Box A. Now, at the start of the game, Box A has a 1/22 chance of containing £250,000. However, you then have to eliminate one box (and, as we are dealing with a situation where you are left at the end with £250,000 and an unknown, then ‘eliminate’ in this case mean getting rid of a non-£250,000 box). At this point the odds that Box A contains the £250,000 goes down to 1/21, because one of the choices has been eliminated. In other words, the scenario now is the same as if you were beginning a game with only 21 boxes. At the second choice, with another box eliminated, the chance that Box A contains the £250,000 goes down to 1/20. And so on, until, when you only have two boxes left, the chance that Box A contains the £250,000 is 1/2. In other words, this is like having a game where you have to choose just between two boxes. Hence swapping makes no difference to your chances.
Conclusion: It’s wrong to say that at the end of the game your box still has a 1/22 chance of containing £250,000. This fails to take into account the fact that odds change when information becomes known. In this case, information becoming known (opening a box) reduces odds by reducing the choices available.
I’d be happy for anyone to point out flaws with this.
Personally I don’t know which is true. But I haven’t seen it described this way here yet.
If we class the 250k as a “win” box, and ALL the rest as “lose” boxes, there are 3 ways a game of DOND can go, assuming we end with 2-box.
1) You choose the “win” box, open 20 other boxes. Left with the “win” and a “lose”.
2) You choose a “lose” box, open 20 other boxes including the “win” and are left with 2 “lose” boxes.
3) You choose a “lose” box, open 20 other boxes but not the “win” and are left with the “win” and the “lose” again.
BUT, and this is a HUUUGEE BUT, we’re not considering ALL possible games here. We’re only considering the ones that end with the “win” box in play.
Therefore you can COMPLETELY IGNORE the probability of having the “win” box left at the end of the game.
Therefore in this subset of games, there are only 2 ways the game can go:
1) You chose the “win” at the start and a “lose” is in the other box.
2) You chose a “lose” at the start and the “win” is in the other box.
These chances are clearly 1/22 and 21/22.
HOWEVER!
The information we have at the end of the game is not just that we have a “win” and a “lose” left.
This is where DOND differs from the MH problem. The “lose” boxes are NOT THE SAME.
We have some extra information.
This means, that out of all the 21 situations where we chose a “lose” and ended up with the “win” in the other box, this is reduced to just 1!
Therefore we’re left with two 1/22 chances, and thus it’s 50:50. Swapping makes no difference.
Where the analogy goes wrong, is where you simplify the problem to being 21 “lose” boxes, and 1 “win” box. If in DOND, all the boxes except the 250k were blank, then there should be a bonus to swapping, as in MH. But since we know which of the 21 “lose” boxes in particular we ended up with, there is no other options.
I think.
OP is right. Difference is simple, in MH problem you actually still have 3 doors to choose from. The door your shown isn’t a choice – you can never lose, you didn’t make this choice you were simply shown what was behind a door. In DOND when you’re down to 3 boxes, you still have that 33% chance of losing the “winning” case before you get to two cases.
Its easier to consider the chance that you are wrong: in MH you have a 66% chance of being wrong initially. When MH shows you that information, you know that 66% of the time you were going to be wrong. That 66% is now represented by the one remaining unopened box.
In the same scenario but with DOND you also have a 66% chance of being wrong. Unfortunately you must now guess again, and have a 33% chance THIS guess will be wrong as well – the host doesn’t spot you a freebie. So you guess randomly, 33% of the time you lose and open the 1 mil w/ 3 cases to go.
If u guess correctly u had a 1/3 chance you were initially right, 2/3 chance you were wrong (-1/3 from your latest guess) so 1/3 chance u were right 1/3 chance u were wrong. However only two cases left, so 1/2 right or wrong. AKA 50% chance whether u swap or not.